katie__michelle

2 years ago

By using substitution:
15x+3y=5
5x+y=-6
Show me how to do this.

Let's manipulate the second equation, we'll get it in terms of y=, and then we'll PLUG it into the first equation.
\[\large 5x+y=-6\]Subtracting 5x from both sides gives us,\[\large y=\color{cadetblue}{-6-5x}\]We want to plug this in for the y in the other equation.\[\large 15x+3\color{cadetblue}{y}=5\]
Which gives us,\[\huge 15x+3\color{cadetblue}{(-6-5x)}=5\]